Answer: (iii) $d = \frac{2}{\|w\|}$

Explanation

For a linear SVM with separating hyperplane

$$w \cdot x + b = 0$$

the perpendicular distance from the origin to the plane is $|b|/\|w\|$. If the margin is defined by the two boundary hyperplanes

$$w \cdot x + b = +1 \quad \text{and} \quad w \cdot x + b = -1$$

then the total margin (width) is the distance between them:

$$d = \frac{|1 - (-1)|}{\|w\|} = \frac{2}{\|w\|}$$

This is the quantity maximised by SVM: maximising the margin $2/\|w\|$ is equivalent to minimising $\frac12\|w\|^2$ subject to the classification constraints.

Why the others are wrong:

• (i) $1/|w|$ — that is the distance of one of the boundary planes from the hyperplane centre (the half-margin), not the full margin.
• (ii) $1/(2|w|)$ — half of option (i), not the margin.
• (iv) $1/|w|^2$ — has no such role in the SVM margin formula.

Hence the margin of the decision boundary in SVM is $d = 2/\|w\|$.